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Some quick calc questions

  • Thread starter don23
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  • #1
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ok, first what is e^t times e^-t?
and can anyone help with this one:

the integral from 0 to pi of (sinti+costj+tk)?
is it -costi + sintj +(t^2/2)k?

thank you!
 

Answers and Replies

  • #2
137
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don23 said:
ok, first what is e^t times e^-t?
and can anyone help with this one:

the integral from 0 to pi of (sinti+costj+tk)?
is it -costi + sintj +(t^2/2)k?

thank you!
Well e^-t x e^t is same as e^t-t so It should be e^0=1.

I havent started on Integrals yet.
 
  • #3
759
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I believe

[tex] e^t e^{ - t} = e^0 = 1 [/tex]

and

[tex]\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2}[/tex]
:smile: Just remember to take the limits of the integral.
You have antiderivative, now apply the algebra
 
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  • #4
VietDao29
Homework Helper
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[tex]\int\limits_0^\pi {\left( {i\sin t + j\cos t + kt} \right)dt} = 2i + k\frac{{\pi ^2 }}{2} + C [/tex]
bomba923, I think you've made a mistake here, :wink:
First, I think the question reads:
[tex]\int_0 ^ \pi (\sin (it) + \cos (jt) + kt) dt[/tex]
And second, where does the little 'C' come from? :smile:
-------------------
don23 said:
is it -costi + sintj +(t^2/2)k?
Nope, it isn't.
In your post, you didn't state the variable of integration. So I assume it is dt.
So:
[tex]\int (\sin (it) + \cos (jt) + kt) dt = \int \sin (it) dt + \int \cos (jt) dt + \int kt dt[/tex]
[tex]= \int \frac{1}{i} \sin (it) d(it) + \int \frac{1}{j} \cos (jt) d(jt) + k \int t dt[/tex]
[tex]= \frac{1}{i} \int \sin (it) d(it) + \frac{1}{j} \int \cos (jt) d(jt) + k \int t dt = ...[/tex]
Can you go from here?
Viet Dao,
 
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  • #5
5
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VietDao29 said:
First, I think the question reads:
[tex]\int_0 ^ \pi (\sin (it) + \cos (jt) + kt) dt[/tex]
You have confused me here.

I thought the question asked:
[tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]

My answer is
[tex] i + j + \frac{{\pi ^2 }}{2}k[/tex]
 
  • #6
VietDao29
Homework Helper
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Buzzer said:
You have confused me here.

I thought the question asked:
[tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]
If I am asking some question like that, I am going to put the i, and j in front of the cos and sin functions.
Something like this:
i sint + j cost + kt.

Buzzer said:
My answer is
[tex] i + j + \frac{{\pi ^2 }}{2}k[/tex]
And even if the question reads like this:
[tex]\int_0 ^ \pi ((\sin t)i + (\cos t)j + (t)k) dt[/tex]
That is still not the correct answer. :wink:
--------------
EDIT:
If the question happens to be:
int (i sint + j cost + kt) dt
You can use bomba923's suggestion. You should omit the constant of integration 'C' in his answer.
Viet Dao,
 
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  • #7
5
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VietDao29 said:
If I am asking some question like that, I am going to put the i, and j in front of the cos and sin functions.
Something like this:
i sint + j cost + kt.
Ok, but I learnt to put i, j and k at the end when I did vectors....

I did it again and got the same answer. Where does 2i come from?


EDIT
How stupid of me, I had pi/2 in my head instead of pi.
 
Last edited:
  • #8
HallsofIvy
Science Advisor
Homework Helper
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Certainly anyone taking calculus should have learned long ago that axbx= ax+y. In particular, axa-x= a0= 1. The fact that a= e is irrelevant.

Yes, the original post's "integral from 0 to pi of (sinti+costj+tk)" should have been written
[tex]\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k} dt[/itex]
to distinguish it form [tex]\int_0^{\pi}{sin ti+ cos tj+ tk} dt[/itex] but I think the "i, j, k" suggested vectors rather than numbers.

don23, It is true that
[tex]\int {(sin t)i+ (cos t)j+ (t)k}dt= -cos(t)i+ sin(t)j+ (\frac{1}{2}t^2+ C[/tex]
but you forgot to evaluate it between 0 and [itex]\pi[/itex].

[tex](-cos \pi)i+ (sin /pi)j+ \frac{1}{2}(/pi)^2= i+ 0j+ \frac{\pi^2}{2}k[/tex]
and
[tex](-cos 0)i+ (sin 0)i+ \frac{1}{2}(0)^2= -i+ 0j+ 0k[/tex]
so
[tex]\int_0^{\pi}{(sin t)i+ (cos t)j+ (t)k}dt= 2i +\frac{\pi^2}{2}k[/tex]
 
  • #9
759
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Thank you Halls of Ivy :redface:
(you've vindicated me!)
 

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